If a , b , c , d and p are different real numbers such that: a 2 + b 2 + c 2 p 2 - 2 a b + b c + c d p + b 2 + c 2 + d 2 ≤ 0 , then show that a , b , c and d are in G.P.

Given that a 2 + b 2 + c 2 p 2 - 2 a b + b c + c d p + b 2 + c 2 + d 2 ≤ 0 a 2 p 2 - 2 a b p + b 2 + b 2 p 2 - 2 b c p + c 2 + ( c 2 p 2 - 2 c d p + d 2 ) ≤ 0 a p - b 2 + b p - c 2 + c p - d 2 ≤ 0 a p - b 2 + b p - c 2 + c p - d 2 ≤ 0 This is possible only when a p - b = 0 ⇒ p = b a b p - c = 0 ⇒ p = c b c p - d = 0 ⇒ p = d c Thus, b a = c b = d c The terms of the G.P. are in the form of a , a r , a r 2 , a r 3 , . . . a r n - 1 are called as a 1 , a 2 , a 3 , a 4 , . . . , a n respectively. Common ratio r = a 2 a 1 = a 3 a 2 = . . . = a n a n - 1 Hence, a , b , c and d are in G.P.

If a + b x a - x = b + c x b - c x = c + d x c - d x , with x ≠ 0 , then show that a , b , c and d are in G.P.

Given that a + b x a - b x = b + c x b - c x = c + d x c - d x a + b x a - b x = b + c x b - c x Applying componendo and dividendo a + b x + a - b x a + b x - a + b x = b + c x + b - c x b + c x - b + c x 2 a 2 b x = 2 b 2 c x a b = b c … i b + c x b - c x = c + d x c - d x Applying componendo and dividendo b + c x + b - c x b + c x - b + c x = c + d x + c - d x c + d x - c + d x 2 b 2 c x = 2 c 2 d x b c = c d … i i From e q n i and e q n i i a b = b c = c d The terms of the G.P. are in the form of a , a r , a r 2 , a r 3 , . . . a r n - 1 are called as a 1 , a 2 , a 3 , a 4 , . . . , a n respectively. Common ratio r = a 2 a 1 = a 3 a 2 = . . . = a n a n - 1 Hence, a , b , c and d are in G.P.